Question: You have found the following ages (in years) of 5 bears. Those bears were randomly selected from the 33 bears at your local zoo: $ 6,\enspace 4,\enspace 25,\enspace 35,\enspace 33$ Based on your sample, what is the average age of the bears? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 33 bears, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{6 + 4 + 25 + 35 + 33}{{5}} = {20.6\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {213.16} + {275.56} + {19.36} + {207.36} + {153.76}} {{5 - 1}} $ {s^2} = \dfrac{{869.2}}{{4}} = {217.3\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{217.3\text{ years}^2}} = {14.7\text{ years}} $ We can estimate that the average bear at the zoo is 20.6 years old. There is also a standard deviation of 14.7 years.